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3 years ago

What is the decimal for 2/3

Mathematics

1 answer:

lakkis [162]3 years ago

7 0

0.6 repeating (AKA 0.6666666-----)

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I NEED HELP!!!!!!!!!!!!

Alex Ar [27]

Answer:

Step-by-step explanation:

123

3 0

3 years ago

What is the inverse of the function of (x)=2^(x)+6

Elden [556K]

Answer:

Option D.

Step-by-step explanation:

To find the inversion of a function follow the following procedure

1) Replace x with y in the function and clear y

$y=2^{(x)} + 6$ ------> $x = 2^{(y)}+ 6$

$x-6 = 2^y$

$y = log_2(x-6)$

2) Check. The range of f(x) is the domain of $f^{-1}(x)$ .

So if f(a) = b, this means that $f^{-1}(b)$ = a.

$f(2) = 2^2+ 6 = 10$

$f^{-1}(x) = log_2(10-6) = 2$

Is fulfilled. Therefore $y=log_2(x-6)$ is the inverse of $f(x) = 2^{(x)} +6$

7 0

3 years ago

Find the equation of the line perpendicular to y = - 4x + 8 that passes through the point (8,4).y =

Kryger [21]

Answer:

y= (1/4)x+2

Explanation:

Comparing y =-4x + 8 to the slope-intercept form y=mx+b

• Slope = -4.

Let the slope of the new line = m

Two lines are perpendicular if the product of their slopes is -1.

Therefore:

$\begin{gathered} -4m=-1 \\ m=\frac{-1}{-4} \\ m=\frac{1}{4} \end{gathered}$

The slope of the perpendicular line = 1/4

Since the line passes through the point (8,4), we have:

$\begin{gathered} y-y_1=m(x-x_1) \\ y-4=\frac{1}{4}(x-8) \\ y=\frac{1}{4}(x-8)+4 \\ y=\frac{1}{4}x-2+4 \\ y=\frac{1}{4}x+2 \end{gathered}$

The equation of the perpendicular line is y= (1/4)x+2

3 0

2 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$